If a parallelogram have the same base with a triangle and be in the same parallels, the parallelogram is double of the triangle.

For let the parallelogram $ABCD$ have the same base $BC$ with the triangle $EBC$, and let it be in the same parallels $BC$, $AE$;I say that the parallelogram $ABCD$ is double of the triangle $BEC$.For let $AC$ be joined.Then the triangle $ABC$ is equal to the triangle $EBC$;for it is on the same base $BC$ with it and in the same parallels $BC$, $AE$. [Prop. 1.37]But the parallelogram $ABCD$ is double of the triangle $ABC$;for the diameter $AC$ bisects it; [Prop. 1.34]so that the parallelogram $ABCD$ is also double of the triangle $EBC$.Therefore etc.Q.E.D.